SOLUTIONS 2. PRACTICE EXAM 2. HOURLY. Problem 1) TF questions (20 points) Circle the correct letter. No justifications are needed.

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1 SOLUIONS 2. PRACICE EXAM 2. HOURLY Math 21a, S03 Problem 1) questions (20 points) Circle the correct letter. No justifications are needed. A function f(x, y) on the plane for which the absolute minimum and the absolute maximum are the same must be constant. rue. Remark. his would not be true if absolute would be replaced by local. he functions f(x, y) and g(x, y) = f(x, y) do not have the same critical points. alse. Because the gradients of f and g agree, also their critical points agree. he sign of the Lagrange multiplier tells whether the critical point of f(x, y) constrained to g(x, y) = 0 is a local maximum or a local minimum. alse. We would get the same Lagrange equations when replacing g with g and λ with λ. he gradient of a function f(x, y, z) is tangent to the level surfaces of f alse. he gradient is normal to the level surface. he point (0, 1) is a local minimum of the function x 3 +(sin(y 1)) 2. alse. While the gradient is (3x 2, 2 sin(y 1) cos(y 1)), the critical point is not a minimum. or any curve, the acceleration vector r (t) of r(t) is orthogonal to the velocity vector at r(t). alse. ake r(t) = (t 2, t). he velocity is (2t, 1), the acceleration (2, 0). heir dot product is 4t. If D u f(x, y, z) = 0 for all unit vectors u, then (x, y, z) is a critical point. rue. If (x, y, z) is not a critical point, then the gradient vector n = f(x, y, z) would have positive length and taking u = n/ n would give D u f(x, y, z) = n 2 0. b a d c x dxdy = (d2 c 2 )(b a)/2, where a, b, c, d are constants. rue. Yes, by direct integration. he functions f(x, y) and g(x, y) = (f(x, y)) 2 have the same critical points. alse. he gradient of g is 2f f. So, the second function has critical points, where f vanishes. If a function f(x, y) = ax + by has a critical point, then f(x, y) = 0 for all (x, y).

2 rue. At a critial point the gradient is (a, b) = (0, 0), which implies f = 0. f xyxyx = f yyxxx for f(x, y) = sin(cos(y + x 14 ) + cos(x)). rue. ollows from Clairot s theorem. he function f(x, y) = x 2002 y 2002 has a critical point at (0, 0) which is a local minimum. alse. It is a local maximum. It is possible that for some unit vector u, the directional derivative D u f(x, y) is zero even though the gradient f(x, y) is nonzero. rue. his happens at a saddle point. If (x 0, y 0 ) is the maximum of f(x, y) on the disc x 2 + y 2 1 then x y 2 0 < 1. alse. he maximum could be on the boundary. he linear approximation L(x, y, z) of the function f(x, y, z) = 3x+ 5y 7z at (0, 0, 0) satisfies L(x, y, z) = f(x, y, z). rue. f(0, 0, 0) = 0 and f(0, 0, 0) = (3, 5, 7). If f(x, y) = sin(x) + sin(y), then 2 D u f(x, y) 2. rue. D u f f 2. here are no functions f(x, y) for which every point on the unit circle is a critical point. alse. here are many rotationally symmetric functions with this property. An absolute maximum (x 0, y 0 ) of f(x, y) is also an absolute maximum of f(x, y) constrained to a curve g(x, y) = c that goes through the point (x 0, y 0 ). rue. he Lagrange multiplier vanishes in this case. If f(x, y) has two local maxima on the plane, then f must have a local minimum on the plane. alse. Look at a camel type surface. It has a saddle between the local maxima. D f(x, y)g(x, y) da = ( D f(x, y) da)( D g(x, y) da) is true for all functions f and g. alse. Example f(x, y) = x 2, g(x, y) = x 3 and where D is the unit square. 2

3 Problem 2) (10 points) Match the parametric surfaces with their parameterization. No justification is needed. I II III IV Enter I,II,III,IV here Parameterization IV (u, v) (u, v, u + v) I (u, v) (u, v, sin(uv)) II (u, v) (0.2 + u(1 u 2 )) cos(v), (0.2 + u(1 u 2 )) sin(v), u) III (u, v) (u 3, (u v) 2, v) Surface I is a graph. Surface II is a surface of revolution. Surface III is algebraic. One of the traces is (u 3, u 2 ), an other trace is the parabola (v 2, v). Surface IV is a plane. 3

4 Problem 3) (10 points) Match the integrals with those obtained by changing the order of integration. No justifications are needed. Enter I,II,III,IV or V here. V I II III Integral y f(x, y) dxdy y f(x, y) dxdy 1 1 y 0 0 f(x, y) dxdy f(x, y) dxdy 1 y 0 0 I) 1 x 0 0 f(x, y) dydx II) 1 1 x 0 0 f(x, y) dydx III) x f(x, y) dydx IV) 1 x f(x, y) dydx V) x f(x, y) dydx Problem 4) (10 points) Consider the graph of the function h(x, y) = e 3x y ind a function g(x, y, z) of three variables such that this surface is the level set of g. 2. ind a vector normal to the tangent plane of this surface at (x, y, z). 3. Is this tangent plane ever horizontal? Why or why not? 4. Give an equation for the tangent plane at (0, 0). 4

5 Solution. 1. g(x, y, z) = e 3x y + 4 z. 2. g(x, y, z) = (3e 3x 0 y 0, e 3x 0 y 0, 1). At the point (x 0, y 0, z 0 ), we have the gradient (a, b, c) = ( 3e 3x 0 y 0, e 3x 0 y 0, 1) and so the plane ax + by + cz = d, where d = ax 0 + by 0 + cz Horizontal would mean a = b = 0 which is not possible because e 3x 0 y 0 is always negative. 4. he tangent plane which goes through the point (0, 0, h(0, 0)) = (0, 0, 5) = (x 0, y 0, z 0 ) is 3x y z = d, where d = = 5. 3x + y + z = 5. Problem 5) (10 points) ind all the critical points of the function f(x, y) = x2 + 3y2 2 2 xy3. or each, specify if it is a local maximum, a local minimum or a saddle point and briefly show how you know. Solution. f(x, y) = x y 3, 3y 3xy 2. his is zero if 3y 3y 5 = 0 or y(1 y 4 ) = 0 which means y = 0 or y = ±1. In the case y = 0, we have x = 0. In the case y = 1, we have x = 1, in the case y = 1, we have x = 1. he critical points are (0, 0), (1, 1), ( 1, 1). he discriminant is f xx f yy f 2 xy = 3 9y 4. he entry f xx is 1 everywhere. Applying the second derivative test gives Critical point (0,0) (1,1) (-1,-1) Discriminant f xx Anslysis min saddle saddle Problem 6) (10 points) Minimize the function E(x, y, z) = k2 8m ( 1 x y z 2 ) under the constraint xyz = 8, where k 2 and m are constants. Remark. In quantum mechanics, E is the ground state energy of a particle in a box with dimensions x, y, z. he constant k is usually denoted by h and called the Planck constant. Solution. Write C = k 2 /(8m) to save typing. E(x, y, z) = 2C(1/x 3, 1/y 3, 1/z 3 ). he constraint is G(x, y, z) = xyz 8 = 0. We have G(x, y, z) = (yz, xz, xy). he Lagrange equations are 2C = λx 3 yz 5

6 2C = λxy 3 z 2C = λxyz 3 xyz = 8 Eliminating λ gives x 2 = y 2 = z 2 and x = y = z = 2 and the minimal energy is 3C/4 = 3k 2 /(32m). Problem 7) (10 points) Assume (x, y) = g(x 2 + y 2 ), where g is a function of one variable. ind xx (1, 2) + yy (1, 2), given that g (5) = 3 and g (5) = 7. Solution. x = g (x 2 + y 2 )2x. xx (x, y) = g (x 2 + y 2 )4x 2 + g (x 2 + y 2 )2. y = g (x 2 + y 2 )2y. yy (x, y) = g (x 2 + y 2 )4y 2 + g (x 2 + y 2 )2. xx + yy (1, 2) = (2 + 2) = 152. Problem 8) (10 points) Consider the region inside x 2 + y 2 + z 2 = 2 above the surface z = x 2 + y 2. a) Sketch the region. b) ind its volume. Solution. a) he intersection of the two surfaces is a circle of radius 1. he region is the bottom of a paraboloid covered with a spherical cap. b) Use cylindrical coordinates: 2π 1 0 ( 2 r 2 r 2 )rdr = (π/3)(2 r 2 ) 3/2 1 0 π/2 = (π/3)(2 3/2 1) π/4. 6

7 Problem 9) (10 points) Draw the gradient vector field of f(x, y) = xy 2x together with a contour map of f. 4 Solution. By translating y by 2 (use new coordinates (u, v) = (x, y 2)) the function becomes f(u, v) = uv. he contour lines are hyperbolas. Draw them first. he gradient field (x, y) = (y, x 2) is orthogonal to the contour lines

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